๐๏ธ ๐งต Prove You're Not Retarded
Anonymous at Sun, 21 Apr 2024 17:33:10 UTC No. 16139112
Reply with your answers below, if you think you're smarter than a fourth grader.
Bonus:
Correct answer is overwritten if reply has trips.
Anonymous at Sun, 21 Apr 2024 17:36:22 UTC No. 16139114
Anonymous at Sun, 21 Apr 2024 17:41:18 UTC No. 16139124
>>16139114
that's a diamond not a square :)
Anonymous at Sun, 21 Apr 2024 17:46:23 UTC No. 16139131
>>16139120
1. Your unshaded region is not a square
2. Your shaded region is not half
Anonymous at Sun, 21 Apr 2024 17:48:07 UTC No. 16139132
>>16139114
This.
Anonymous at Sun, 21 Apr 2024 17:48:25 UTC No. 16139135
i hate problems like these, they're so simple but I'm too dumb to think outside of the box for solutions like >>16139114
I see the 4 squares as a grid and can immediately only think in terms of an up and down grid
Anonymous at Sun, 21 Apr 2024 17:53:38 UTC No. 16139143
>>16139131
>1. Your unshaded region is not a square
if a square with a square in it is not a square then invert the light/shadow parts
>2. Your shaded region is not half
well..it's clearly low effort
Anonymous at Sun, 21 Apr 2024 18:19:32 UTC No. 16139178
>>16139124
False. It can be proven it has four 90 degree angles because the triangles have equal sides and because parallel lines create equal angles when both cross the same straight line.
Anonymous at Sun, 21 Apr 2024 18:20:20 UTC No. 16139179
>>16139178
All those things are true and it's still a diamond.
Anonymous at Mon, 22 Apr 2024 00:38:27 UTC No. 16139679
>>16139124
it's a square, the orientation doesn't matter.
Anonymous at Mon, 22 Apr 2024 00:39:21 UTC No. 16139681
>>16139621
cool
Anonymous at Mon, 22 Apr 2024 01:38:59 UTC No. 16139751
>>16139112
The answer is "yes." No evidence needed.
>>16139114
>>16139621
>>16139681
Wrong. See above. The question wasn't to fill, it was to answer yes or no.
Anonymous at Mon, 22 Apr 2024 01:42:11 UTC No. 16139754
>>16139112
>Prove You're Not Retarded
But I am retarded.
Anonymous at Mon, 22 Apr 2024 02:06:12 UTC No. 16139776
>>16139751
>The answer is "yes."
Incorrect answer. Forgot to show work.
Anonymous at Mon, 22 Apr 2024 04:19:50 UTC No. 16139863
>>16139145
Haha very funny anon, grow up
Anonymous at Mon, 22 Apr 2024 04:26:37 UTC No. 16139868
>>16139863
>grow up
I gotchu senpai.
Anonymous at Mon, 22 Apr 2024 04:56:58 UTC No. 16139912
Let [math]a[/math] be the side length of the smaller squares. Then the greater square has area [math]4a^2[/math], and we must construct a square with area [math]2a^2[/math]. Such a square would have side length [math]\sqrt{2}a[/math]. By the Pythagorean Theorem, this length can be constructed as the hypotenuse of a right triangle whose legs have length [math]a[/math]. Create this length by joining two opposite vertices of any of the smaller squares, square this length, and shade the remainder of the larger square that lies outside of this construction. Conclude by realizing that you're retarded and have overthought this; thus it has been shown.
Anonymous at Mon, 22 Apr 2024 11:35:27 UTC No. 16140338
>>16139912
>Pythagorean Theorem,
Heresy.
Anonymous at Mon, 22 Apr 2024 11:42:52 UTC No. 16140343
>>16139863
you posted a grade school geometry question. you dont get to tell anyone to grow up
Anonymous at Mon, 22 Apr 2024 11:43:27 UTC No. 16140344
>>16139114
Every piece is a triangle, and a triangle is a square.
Anonymous at Mon, 22 Apr 2024 12:35:24 UTC No. 16140397
>>16139681
That's less than half.
Anonymous at Mon, 22 Apr 2024 15:20:31 UTC No. 16140591
>>16139112
(1-x)^2 = 2(x*(1-x)) + x^2
solve for x to find the boxes to shade
>>16139135
>I'm too dumb to think outside of the box
you're supposed to think inside the box in this case
Anonymous at Mon, 22 Apr 2024 17:17:03 UTC No. 16140753
>>16139112
fuck you i won't do what you tell me